| ⇦ | 
| ⇨ |
In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced is 20 V. The inductance L is
Options
(a) 62.5 mH
(b) 625 mH
(c) 6.25 mH
(d) 0.625 mH
Correct Answer:
62.5 mH
Explanation:
Induced e.m.f. |e| = L (dI / dt) = L(18 – 2) / 0.05 or,
L = (20 × 0.05) / 16 = 0.0625 H = 62.5 mH
Related Questions: - A source of unknown frequency gives 4 beats/s, when sounded with a source harmonic
- The core of a transformer is laminated because
- A road of width 20 m forms an arc of radius 15 m,its outer edge is 2m higher than
- The sodium nucleus ₁₁Na²³ contains
- Perfectly black body radiates the energy 18 J/s at 300K. Another ordinary body of e=0.8
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A source of unknown frequency gives 4 beats/s, when sounded with a source harmonic
- The core of a transformer is laminated because
- A road of width 20 m forms an arc of radius 15 m,its outer edge is 2m higher than
- The sodium nucleus ₁₁Na²³ contains
- Perfectly black body radiates the energy 18 J/s at 300K. Another ordinary body of e=0.8
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply